aa r X i v : . [ m a t h . G R ] J a n ON THE FINITE INDEX SUBGROUPS OF HOUGHTON’SGROUPS
CHARLES GARNET COX
Abstract.
Houghton’s groups H , H , . . . have been studied in many con-texts, and various results exist for their finite index subgroups. In this note wedescribe all of the finite index subgroups of each Houghton group, and theirisomorphism types. Using the standard notation that d ( G ) denotes the mini-mal size of generating set for G we then show, for each n ∈ { , , . . . } and U of finite index in H n , that d ( U ) ∈ { d ( H n ) , d ( H n ) + 1 } and characterise wheneach of these cases occurs. Introduction
Introduced in [10] by Houghton, the Houghton groups have since attracted atten-tion for their finiteness properties [2, 11], their growth [3, 13], their many interestingcombinatorial features [12, 1, 9, 14] as well as other properties.
Notation.
For X = ∅ , let Sym( X ) denote the group of all bijections on X . For g ∈ Sym( X ), let supp( g ) := { x ∈ X : ( x ) g = x } , called the support of g .Then FSym( X ) Sym( X ) consists of elements with finite support, and FSym( X )contains Alt( X ) as a unique index 2 subgroup consisting of the even permutations.We give a brief overview of these groups for our purposes; more detailed intro-ductions can be found, for example, in [1, 4]. Definition.
For any n ∈ { , , . . . } the nth Houghton group , denoted H n , is gener-ated by g , . . . , g n ∈ Sym( X n ) where X n = { , . . . , n } × N and for k ∈ { , . . . , n } (1) ( i, m ) g k = (1 , m + 1) if i = 1 and m ∈ N (1 ,
1) if i = k and m = 1( k, m −
1) if i = k and m ∈ { , , . . . } ( i, m ) otherwise . We will use right actions throughout and define N := { , , . . . } . Notation.
Let G f H denote that G is a finite index subgroup of H . Notation.
For a group G and g, h ∈ G , let [ g, h ] := g − h − gh and g h := h − gh .Observe, for n ∈ { , , . . . } , that FSym( X n ) H n . This can be seen by con-sidering [ g , g ]. We also have a surjective map π : H n → Z n − , π ( g i ) := e i − for i = 2 , . . . , n , where e i denotes the vector in Z n − with i th entry 1 and other entries0. For n ∈ { , , . . . } , this provides a short exact sequence, as observed in [14].1 −→ FSym( X n ) −→ H n π −→ Z n − −→ Date : January 29, 2021.2010
Mathematics Subject Classification.
Key words and phrases. generation, generation of finite index subgroups, structure of finiteindex subgroups, infinite groups, Houghton groups, permutation groups, highly transitive groups.
Definition.
The second Houghton group is generated by the two cycle ((1 , , (1 , g , defined analogously to (1) above. This is isomorphicto FSym( Z ) ⋊ h t i where t ∈ Sym( Z ) sends each z ∈ Z to z + 1.There have been many papers working with the finite index subgroups of thisfamily of groups, e.g. [14, 8, 3, 4, 5, 6]. We start by giving the first full descriptionof them. Theorem A.
Let n ∈ { , , . . . } and U f H n . Then U is either: i) equal to h FSym( X n ) , g c i i : i = 2 , . . . , n i ; or ii) isomorphic to h Alt( X n ) , g c i i : i = 2 , . . . , n i for some constants c , . . . , c n ∈ N . If U f H , then either (i) or (ii) above occursfor some c ∈ N or it is equal to h Alt( X ) , ((1 , , (1 , g ) i . As a result of our methods, we also obtain the following.
Corollary A.
Let n ∈ { , , . . . } . If U f H n , then there exist c , . . . , c n ∈ N suchthat h c e , . . . , c n e n − i = π ( U ) . Then, either: • at least two of c , . . . , c n are odd and U = h g c , . . . , g c n n , FSym( X n ) i ; or • there are n − + 1 distinct possibilities for U . We then extend the work of [7], where the groups h Alt( X ) , g c i were shown tobe 2-generated for each c ∈ N , by investigating the generation properties of each ofthese groups. Notation.
For a finitely generated group G , let d ( G ) := min {| S | : h S i = G } .It is a fact generally seen in an undergraduate group theory course that a finiteindex subgroup of a finitely generated group is itself finitely generated. In the caseof Z n , where n ∈ N , we have that each A f Z n satisfies d ( A ) = d ( Z n ). Our firsttheorem states that this is almost the case for the Houghton groups, which have d ( H ) = 2 and d ( H n ) = n − n ∈ { , , . . . } . We can further categorise exactlywhen d ( U ) = d ( H n ) and when d ( U ) = d ( H n ) + 1. Theorem B.
Let n ∈ { , . . . } and U f H n . Then d ( U ) ∈ { d ( H n ) , d ( H n ) + 1 } . Our methods again provide an interesting corollary.
Corollary B.
Let n ∈ { , , . . . } and G := h Alt( X n ) , g , . . . , g n i f H n . If U f G , then d ( U ) = d ( G ) = d ( H n ) . Similarly if U f H , then d ( U ) = d ( H ) . The structure of finite index subgroups of H n In this section we prove Theorem A. The following is well known.
Lemma 2.1.
Given U f G there exists N f U which is normal in G . Some structure regarding finite index subgroups of H n is known; the followingis particularly useful to us. Lemma 2.2. [5, Prop. 2.5]
Let X be a non-empty set and Alt( X ) G Sym( X ) .Then G has Alt( X ) as a unique minimal normal subgroup. Let n ∈ { , , . . . } and U f H n . By Lemma 2.1, U contains a normal subgroupof H n and so, by Lemma 2.2, Alt( X n ) U . Furthermore, U f H n and so π ( U ) f π ( H n ) where π : H n → Z n − sends g i to e i − for i = 2 , . . . , n . Hence N THE FINITE INDEX SUBGROUPS OF HOUGHTON’S GROUPS 3 there exist minimal k , . . . , k n ∈ N such that π ( g k i i ) ∈ π ( U ) for each i ∈ { , . . . , n } .But π ( g k i i ) = k i e i − , and so the preimage of π ( g k i i ) is { σg k i i : σ ∈ FSym( X n ) } .Together with the fact that Alt( X n ) U , we have(2) U = h g k , . . . , g k n n , FSym( X n ) i or U = h ǫ g k , . . . , ǫ n g k n n , Alt( X n ) i where each ǫ i is either trivial or ((1 ,
1) (1 , H n where n ∈ { , , . . . } . Notation.
For any given n ∈ { , , . . . } and any c = ( c , . . . , c n ) ∈ N n − , let F c := h FSym( X n ) , g c , . . . , g c n n i and G c := h Alt( X n ) , g c , . . . , g c n n i .It is worth noting that there are some c ∈ N n such that G c = F c . Notation.
Let I n := { c = ( c , . . . , c n ) ∈ N n : G c = F c } .We describe I n in Lemma 3.8. This, together with (2), yields Corollary A. Notation.
For n ∈ { , , . . . } and i ∈ { , . . . , n } , let R i := { ( i, m ) : m ∈ N } ⊂ X n . Proposition 2.3.
Let n ∈ { , , . . . } , k = ( k , . . . , k n ) ∈ I n , each ǫ , . . . , ǫ n beeither trivial or (( i,
1) ( i, , and U = h Alt( X n ) , ǫ i g k i i : i = 2 , . . . , n, i f H n .Then U is isomorphic to G k .Proof. If there are i = j such that c i = c j = 1, then FSym( X n ) G c and so G c = F c i.e. c I n .Now consider if c i = 1 for some i and that ǫ i = (( i,
1) ( i, ǫ i g i fixes( i, i, m ) ǫ i g i = ( i, m −
1) for all m >
3, ( i, ǫ i g i = (1 , , m ) ǫ i = (1 , m + 1) for all m >
1, and that ǫ i g i fixes all other points in X n .This allows us to relabel X n so that the point ( i,
1) becomes part of another ray j ∈ { , . . . , n } .If c i = 1 and ǫ i = (( i,
1) ( i, R i as R ′ i by swapping thelabels on the points ( i,
1) and ( i, ǫ i g c i i acts on R ′ i ∪ R in the same wayas g c i i acts on R i ∪ R . (cid:3) Lemma 2.4.
Let k ∈ { , , . . . } , ǫ = ((1 ,
1) (1 , , and U = h Alt( X n ) , ǫg k i f H .Then U is isomorphic to h Alt( X ) , g k i .Proof. In a similar way to the preceding proof, we can relabel R as R ′ (by swappingthe labels on the points (2 ,
1) and (2 , ǫg k acts on R ∪ R ′ in the sameway that g k acts on R ∪ R . (cid:3) Remark 2.5.
The preceding lemma may hold for k = 1 . Such an isomorphismwould not be induced by a permutation of X however, since any element g in h Alt( X ) , ((1 ,
1) (1 , t i with π ( g ) = 1 cannot have an infinite orbit equal to X . Generation of the groups F c and G c We start with the case of H , first dealing with the ‘exceptional’ case. Lemma 3.1.
The group h Alt( X ) , ((1 ,
1) (1 , g i is -generated.Proof. Note that h Alt( X ) , ((1 ,
1) (1 , g i ∼ = h Alt( Z ) , (1 2) t i FSym( Z ) ⋊ h t i where t : z → z + 1 for all z ∈ Z . Our aim will be to show that S = { (0 1 2) , (0 1) t } generates h Alt( Z ) , (1 2) t i . To do this we will use h S i to construct all 3-cycles (0 a b ) CHARLES GARNET COX with 0 < a < b . With these elements we can then also produce every 3-cycle ofthe form ( a b c ) where 0 < a < b < c , by conjugating (0 a b ) by (0 c c ). Thenevery 3-cycle in Alt( N ∪ { } ), or its inverse, is accounted for by such elements.Conjugation by some power of (0 1) t yields every 3-cycle in Alt( Z ), and we recallthat Alt( Z ) is generated by the set of all 3-cycles with support in Z .In addition to our first simplification, note that it is sufficient to show that(0 , , k + 1) ∈ h (0 1 2) , (0 1) t i for every k ∈ N . This is because any 3-cycle (0 a b )with 0 < a < b will be conjugate, by (0 1) t , to an element of the form (0 , , k + 1).We start with σ := (0 , k, k + 1). If k = 1, then we are done. Otherwise, conjugate σ − by (0 , k − , k ) to obtain σ = (0 , k − , k + 1). If k >
2, conjugate σ − by(0 , k − , k −
1) to obtain σ . Continuing in this way yields the result. (cid:3) The following results conclude the H case. We include the proof here as we willadapt it for F c and G c . The notation Ω k := { , . . . , k } is helpful for these results. Lemma 3.2. [7, Lem. 3.6]
Let k ∈ { , , . . . } and t : z → z + 1 for all z ∈ Z . Then h Alt( Z ) , t k i is generated by Alt(Ω k ) ∪ { t k } . Lemma 3.3. [7, Lem. 3.7]
Let k ∈ N and t : z → z + 1 for all z ∈ Z . Then G k := h Alt( Z ) , t k i and F k := h FSym( Z ) , t k i are 2-generated.Proof. We start with G k . We will show that we can find, for each k ∈ { , , . . . } an α k ∈ Alt( Z ) such that h t k , α k i contains Alt( Z ). Then, for d ∈ { , , } , h t d , α i contains h t , α i and so contains Alt( Z ). We can therefore fix some k ∈ { , , . . . } ,meaning all 3-cycles in Alt(Ω k ) are conjugate.Let r = (cid:0) k (cid:1) and let ω , . . . , ω r be a choice of distinct 3-cycles in Alt(Ω k )with ω i = ω − j for every i, j ∈ { , . . . , r } . Thus h ω , . . . , ω r i = Alt(Ω k ). Set σ = (1 3) and σ r +1 = (2 3) and note that ( σ − r +1 σ − σ r +1 ) σ = (1 2 3). Now choose σ , . . . , σ r ∈ Alt(Ω k ) so that for each m ∈ { , . . . , r } we have σ − m (1 2 3) σ m = ω m .Let α k := Q r +1 i =0 t − ik σ i t ik and, for m ∈ { , . . . , r + 1 } , let β m := t mk α k t − mk .Then β − r +1 α − k β r +1 α k = σ − r +1 σ − σ r +1 σ = (1 2 3) and, for m ∈ { , . . . , r } , we havethat β − m (1 2 3) β m = ω m . Hence h α k , t k i = G k .We can now adapt the element α k to an element γ k so, for each k ∈ { , , . . . } ,we have that h γ k , t k i = F k . One way to do so if to set γ k := ( σ ′ )( Q r +1 i =1 t − ik σ i t ik )where σ ′ = (1 3)(4 5). This means that [ σ r +1 , σ ′ ] still equals (1 2 3), and soAlt( Z ) h γ k , t k i , but that Alt( Z ) ∪ γ k (Alt( Z )) = FSym( Z ) h γ k , t k i as well. (cid:3) We now work with a fixed n ∈ { , , . . . } . The following commutator identitieswill be helpful.(3) [ a, bc ] = a − ( bc ) − a ( bc ) = a − c − acc − a − b − abc = [ a, c ][ a, b ] c (4) [ ab, c ] = ( ab ) − c − ( ab ) c = b − ( a − c − ac ) bb − c − bc = [ a, c ] b [ b, c ] Proposition 3.4.
Given c = ( c , . . . , c n ) ∈ I n , we have that d ( G c ) = d ( H n ) .Proof. By possibly relabelling the branches of X n , set c := max { c , . . . , c n } . Thus c = 1, as then c = c = 1 and c I n . Let k := 2 c and Ω ∗ := { (1 , , . . . , (1 , k ) } .Thus 2 k > ∗ ) are conjugate in Alt(Ω ∗ ). We claim thatAlt( X n ) h Alt(Ω ∗ ) , g c , . . . , g c n n i . First, by Lemma 3.2, h Alt(Ω ∗ ) , g c i is a sub-group of H n which generates h Alt( R ∪ R ) , g c i ∼ = h Alt( Z ) , t c i . Most importantly, h Alt(Ω ∗ ) , g c i contains Alt( R ). Secondly, given any σ ∈ FSym( X n ), there exists N THE FINITE INDEX SUBGROUPS OF HOUGHTON’S GROUPS 5 a word of the form g d . . . g d n n for some d i ∈ c i N which conjugates σ to σ ′ , wheresupp( σ ′ ) ⊂ R . Hence h Alt(Ω ∗ ) , g c , . . . , g c n n i = G c . Our aim is now to adapt thepreceding lemma. In particular, we will show that there exists an α ∈ Alt( X n ) suchthat h αg c , g c , . . . , g c n n i = G c . For any ω ∈ FSym( X n ), let σ ω := [ ωg c , g k ], whichwe can calculate using (4) as(5) [ ω, g k ] g c [ g c , g k ] = g − c ( ω − g − k ωg k ) g c [ g c , g k ] . Observe that supp([ g c , g k ]) ⊆ { (1 , , . . . , (1 , c + 2 k ) } and, since k = 2 c > c ,that (1 , k +1) , (1 , k +3) supp([ g c , g k ]). Let δ := ((1 , k − c +1) , (1 , k − c +3))and note, using (5), that supp( σ δ ) ∩ { (1 , m ) : m = 4 k + 2 or m > k + 4 } = ∅ and σ δ swaps (1 , k + 1) and (1 , k + 3). We now restrict to those ω ∈ Alt( X n ) withsupp( ωδ ) ⊂ [ j ∈ N (Ω ∗ ) g − jk . The following implications of this restriction are all important in relation to (5): • supp( σ ω ) ⊂ R ∪ R ; • supp( ω − ) ∩ supp( g − k ωg k ) = ∅ ; • ( x ) g − c ω − g c = ( x ) ω − for all x ∈ R ; and • ( x ) g − c ( g − k ωg k ) g c = ( x ) g − k ωg k for all x ∈ R .Now, by mimicking the proof of Lemma 3.3, we can choose one such ω , which wewill denote by α , so that Alt( R ) h g k , σ α i . Let r = (cid:0) k (cid:1) and ω , . . . , ω r be a set of3-cycles such that h ω , . . . , ω r i = Alt(Ω ∗ ). Choose σ , . . . , σ r ∈ Alt(Ω ∗ ) such that σ − i ((1 ,
1) (1 ,
2) (1 , σ i = ω i for i = 1 , . . . , r . Also set σ r +1 := ((1 ,
2) (1 , σ := ((1 ,
1) (1 , δ i := g ki σ i g − ki for i = 1 , . . . , r +1 and α := δ . . . δ r +1 .Define β m := g − mk σ α g mk for m = − , . . . , r + 1. Then, as in Lemma 3.3, we havethat [ β r +1 , β − ] = [ σ d +1 , σ ] and β i [ σ d +1 , σ ] β − i = ω i for i = 1 , . . . , r . (cid:3) We now look at the final case of determining d ( F c ). Remark 3.5.
Let a, b, c ∈ H n and sgn denote the sign function on FSym( X n ) .From the identities (3) and (4) above, we have that sgn([ a, bc ]) = sgn([ a, b ])sgn([ a, c ]) and sgn([ ab, c ]) = sgn([ a, c ])sgn([ b, c ]) . The hypothesis in the following is critical to our final arguments.
Corollary 3.6.
Let c = ( c , . . . , c n ) . If there exist i, j ∈ { , . . . , n } such that [ g c i i , g c j j ] ∈ FSym( X n ) \ Alt( X n ) , then d ( F c ) = d ( H n ) .Proof. By the preceding proposition, we have that there exists an α ∈ Alt( X n ) suchthat G c h αg c , g c , . . . , g c n n i . If [ g c i i , g c j j ] ∈ FSym( X n ) \ Alt( X n ) then, by Remark3.5, so is [ αg c i i , g c j j ]. Hence h αg c , g c , . . . , g c n n i ∩ (FSym( X n ) \ Alt( X n )) = ∅ and soequals F c . (cid:3) Our aim is now to prove the converse of the preceding statement.
Proposition 3.7.
Let c = ( c , . . . , c n ) . If for every i, j ∈ { , . . . , n } we have that [ g c i i , g c j j ] ∈ Alt( X n ) , then d ( F c ) = d ( H n ) + 1 .Proof. Recall that d ( G c ) = n −
1. Including an element from FSym( X n ) \ Alt( X n )to the generating set for G c therefore yields a generating set of F c of size n . Wetherefore need only show that, with the given hypothesis, no generating set of F c of size n − S denote such a set. Then the CHARLES GARNET COX homomorphism π : H n → Z n − and the fact that F c is finite index in H n togetherimply that π ( S ) must be linearly independent. Thus S cannot contain any elementsfrom FSym( X n ). Moreover, given a word w in S that represents an element inFSym( X n ), for each s ∈ S the sum of the exponents relating to all occurrencesof s must be zero (since π − ( x ) = FSym( X n ) if and only if x = ). Therefore,by using the proof that Z n − has quadratic Dehn function, w can be rewrittenas a product of conjugates of commutators in S . Let σ Q i ∈ I g p i i , ω Q j ∈ J g q j j ∈ S ,where σ, ω ∈ FSym( X n ). Then their commutator, using (3), will be a productof conjugates of elements of the form [ σ Q i ∈ I g p i i , ω ] or [ σ Q i ∈ I g p i i , g q k k ] for some k ∈ J . Applying (4) will produce a product of conjugates of elements of the form[ σ, ω ] , [ g p l l , ω ] , [ σ, g q k k ] or [ g p l l , g q k k ]for some l ∈ I and k ∈ J . The first 3 of these are clearly in Alt( X n ), and the 4th is inAlt( X n ) by our assumption that [ g c i i , g c j j ] ∈ Alt( X n ) for all i, j ∈ { , . . . , n } . Hence,from Remark 3.5, a commutator of any two elements in S will be in Alt( X n ). (cid:3) We end by describing the set I n . Lemma 3.8.
Let c = ( c , . . . , c n ) ∈ N n . Then G c = F c if and only if there existdistinct i, j ∈ { , . . . , n } such that c i and c j are both odd.Proof. From the preceding two results, G c = F c if and only if there exist i and j in { , . . . , n } such that [ g c i i , g c j j ] ∈ FSym( X n ) \ Alt( X n ). By repeatedly applyingRemark 3.5, we can reduce c i and c j to be either 1 or 2, depending on whetherthey are odd or even respectively. Then [ g i , g j ] , [ g i , g j ] , [ g i , g j ] ∈ Alt( X n ) whereas[ g i , g j ] ∈ FSym( X n ) \ Alt( X n ). (cid:3) References [1] Y. Antol´ın, J. Burillo, and A. Martino,
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